Question

Question 10
$cos\theta =\frac{a^2+b^2}{2ab}$, where a and b are two distinct numbers such that ab>0.

Answer 1

The statement is false.

Given, a and b are two distinct numbers such that ab>0.

Using AM>GM,

[AM and GM of two number a and b are $\frac{a+b}{2}$ and $\sqrt{ab}$, respectively]

$\Rightarrow \frac{a^2+b^2}{2}>\sqrt{a^2.b^2}$

$\Rightarrow a^2+b^2>2ab$

$\Rightarrow \frac{a^2+b^2}{2ab}>1[\because cos\theta =\frac{a^2+b^2}{2ab}]$

$\Rightarrow cos\theta >1[\because -1\le cos\theta \le 1]$

Which is not possible.

Hence, $cos\theta \ne \frac{a^2+b}{2ab}$.